Regression and percentile 
On a midterm exam, the average is $50$ points (out of $100$) with an SD of $10$ points.  On the final exam, the average was $100$ points (out of $200$) with an SD of 30 points.  The correlation between the two is 0.6. 
What percent of the people who got a 40 on the midterm improved their percentile on the final?

The solution given in this case is 31% but it doesn't seem right to me .
I started the question by first standardize the score of 40 on midterm:
$\frac{40-50}{10}=-1$ 
Thus the same percentile of score on final would be $-1*30+100=70$
I then solve for the new average for the final score of the people who got 40 in the midterm, I get: $-1*0.6*30+100=82$
Finally I solve for the z score of the people who were 1 standard deviation below the new average and use normal approximation to get the percentage of people who improved on their final grade:
$\frac{70-82}{\sqrt{1-0.6^2}*30} = -0.5$
-0.5 corresponds to 30.85% under the curve, we are interested in people who improved in their final score therefore we get: 100%-30.85%=69.15% 
Is my solution correct in this case??
 A: I will solve this according to the least-squares regression line equation.
The linear relation between $2$ variables $x$ & $y$ is given by the equation for the regression line,$$ Y= \beta_o+\beta_1Χ+e.....(1)$$
Where $Y$ is the dependent variable
$Χ$ is the independent variabl
$\beta_o$ is the y-intercept.It is the value of Υ for Χ=0
$\beta_1$ is the slope of the line.It gives the amount of change in $Y$ for every unit change in $Χ$.
$e$ is the random error.
The predicted value of $Y$ for a given value of $Χ$ is denoted by $Y^{\prime}$
$$Y^{\prime}=a+bx.....(2)$$where $a= \beta_0^{\prime}=$ the estimated $y$-intercept of the regression line
and $b=\beta_1^{\prime}=$ the estimated slope of the regression line.
Regression line of $y$ on $x$ is $y-\overline{y}=byx(x-\overline{x}).....\overline{y}$ is the mean of $y,\overline{x}$ is the mean of $x.....(3)$
we consider $x$ is the midterm exam score
$y$ is the final exam score
Given midterm mean $=50=\overline{x},\overline{y}=100$
and the standard deviation for $x=10$ and $y=30$ $$byx=r\times\dfrac{\sigma}{\sigma x}=0.6\times\dfrac{30}{10}=1.8$$
Plug in these values in $(3)$, we get
$$y-50=1.8(x-50)$$
$$y=1.8x+10$$$\beta_0$ is the $y$-intercept $=10$
$\beta_1$ is the slope of the line $=1.8$
Now, plug these in $(1)$, we get
If $X=40$ then 
$Y^{\prime}(40)=82$
Therefore, the predicted final exam score is $82$
